The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in

Transcription

1 The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 7 or higher.

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3 Problem C A Time For Change A customer purchased some cough candies for $1.73 and paid for them with a two-dollar coin. The cashier made the correct change using only pennies, nickels, dimes and quarters. In how many different ways can the cashier make change?

4 Problem C and Solution A Time For Change Problem A customer purchased some cough candies for $1.73 and paid for them with a two-dollar coin. The cashier made the correct change using only pennies, nickels, dimes and quarters. In how many different ways can the cashier make change? Solution This is a good problem for applying a systematic approach. The amount of change required is $2 $1.73 = $0.27 or 27 cents. In order to get to 27 a minimum of 2 pennies are required. We can systematically look at all of the possibilities using pennies. The number of pennies given must end in a 2 or a 7 so that when this number is subtracted from 27 the result will be a number ending in 0 or 5. This result can then be achieved using nickels, dimes or quarters. Therefore, we can use 2, 7, 12, 17, 22 or 27 pennies. The following chart presents the possibilities. Number of Pennies Number of Nickels Number of Dimes Number of Quarters there are 13 different ways to make the correct change.

5 Problem C Chocolates or Cards? For Valentine s Day, in a class of 30 students, 12 students brought chocolates for the teacher, 17 brought the teacher a card, and 5 students did both. How many students did not bring the teacher chocolates or a card?

6 Problem C and Solution Chocolates or Cards? Problem For Valentine s Day, in a class of 30 students, 12 students brought chocolates for the teacher, 17 brought the teacher a card, and 5 students did both. How many students did not bring the teacher chocolates or a card? Solution Since 5 students did both and these students are included in the 12 who brought chocolates, then 12 5 or 7 students brought chocolates only. They did not bring a card for the teacher as well. Again, since 5 students did both and these students are included in the 17 who brought the teacher a card, then 17 5 or 12 students brought a card only. They did not bring chocolates for the teacher as well. Students will be in one of four possible groups: they brought both chocolates and a card, they brought chocolates only, they brought a card only, or they did not bring a card or chocolates. The number of students in each group added together will sum to the number of students in the class. Or we could subtract the sizes of the known groups from the class size to get the number of students who did neither. So, the number of students who did neither is the number of students in the class minus the number of students who brought both chocolates and a card minus the number of students who brought chocolates only minus the number of students who brought a card only. Therefore the number of students who did neither is equal to or 6. Six students did not bring their teacher chocolates or a card. These students will have to find some other way to show their fondness for their teacher.

7 Problem C Better Than Average Stu Dent has six marks on his report card. His overall average is 75%. But after looking more closely at his marks, Stu discovered that his mathematics mark was incorrectly recorded as a 39% instead of his actual mark of 93%. Determine Stu Dent s correct report card average.

8 Problem C and Solution Better Than Average Problem Stu Dent has six marks on his report card. His overall average is 75%. But after looking more closely at his marks, Stu discovered that his mathematics mark was incorrectly recorded as a 39% instead of his actual mark of 93%. Determine Stu Dent s correct report card average. Solution To calculate an average we add the six marks and divide by 6. sum of six marks 6 = 75 To then obtain the sum of the six marks we would multiply the average by 6. sum of six marks = 6 75 = 450 But this sum includes the wrong mathematics mark of 39%. So we need to adjust the sum by subtracting the wrong mark and adding the corrected mark. correct sum of six marks = = 504 We can now obtain Stu s corrected average by dividing the corrected sum by 6. correct average = correct sum of six marks 6 = = 84 Stu s corrected average is 84%. His day is now much better than average!

9 Problem C Some Wet Weather Math? Raynor Schein has a bag that contains three black marbles, five gold marbles, two purple marbles, and six red marbles. During the course of cleaning his room one wet spring day, Raynor finds some white marbles and adds them to the bag. Raynor tells his friend April Showers that if she now draws a marble at random from the bag, the probability of it being black or gold is 2 7. How many white marbles did Raynor Schein add to the bag?

10 Problem C and Solutions Some Wet Weather Math? Problem Raynor Schein has a bag that contains three black marbles, five gold marbles, two purple marbles, and six red marbles. During the course of cleaning his room one wet spring day, Raynor finds some white marbles and adds them to the bag. Raynor tells his friend April Showers that if she now draws a marble at random from the bag, the probability of it being black or gold is 2. How many white marbles did Raynor Schein add to the bag? 7 Solution 1 At present there are = 16 marbles in the bag. Since after adding some marbles to the bag the probability of picking a black or gold marble is 2, this implies that the new total 7 number of marbles in the bag is a multiple of 7 and this multiple must be greater than 16. Therefore there are possibly 21, 28, 35, 42, 49, 56, marbles in the bag. There are a total of = 8 black and gold marbles in the bag. If k is the total number of marbles in the bag after adding some white marbles, then 8 k = 2 7 but 2 7 = 8 28 so 8 k = 8 28 and k = 28 follows. Since there were 16 marbles in the bag and there are now 28 marbles in the bag, Raynor added = 12 white marbles to the bag. Therefore, Raynor added 12 white marbles to the bag. Solution 2 Let w be the number of white marbles added to the bag. There are now w = 16 + w marbles in the bag and = 8 black and gold marbles in the bag. We know that the number of black and gold marbles divided by the total number of marbles in the bag is 2 7 so w = 2 7. Cross-multiplying we get (8)(7) = (2)(16 + w) which simplifies to 56 = w and 24 = 2w. The result w = 12 follows. Therefore, Raynor added 12 white marbles to the bag.

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12 Problem C A Bit of Shade Please A B 3 cm O In the diagram, AOB = 90 and the radius of the circle is OA = 3 cm. Determine the area of the shaded region correct to one decimal place.

13 Problem C and Solution A Bit of Shade Please Problem In the diagram, AOB = 90 and the radius of the circle is OA = 3 cm. Determine the area of the shaded region correct to one decimal place. A 3 cm O B Solution The triangle plus the shaded area cover 1 4 of the area of the circle. To find the shaded area we need to find the area of the triangle and subtract it from the area of 1 4 of a circle with radius 3 cm. Since AOB = 90, AO and OB meet at 90, we can use OA as the base and OB as the height in the formula for the area of a triangle, base height 2. Area AOB = (AO) (OB) 2 = (3) (3) (2) = 9 2 = 4.5 cm 2. To find the area of the quarter circle we will use the formula πr 2 and divide the result by 4. Area of Quarter Circle = π r 2 4 = π = π 9 4 = 2.25π cm 2. We can now determine the shaded area. Shaded area = Area of Quarter Circle Area of Triangle = 2.25π 4.5. = 2.6 cm 2. Therefore, the shaded area is approximately 2.6 cm 2.

14 Problem C The Missing Link F A D B C E ABCD is a rectangle inscribed in a quarter-circle as shown. A is on BF, B is the centre of the quarter-circle, C is on BE, and D is on arc F E. If AD = 12 cm and CE = 1 cm, determine the length of AF.

15 Problem C and Solution The Missing Link Problem ABCD is a rectangle inscribed in a quarter-circle as shown. A is on BF, B is the centre of the quarter-circle, C is on BE, and D is on arc F E. If AD = 12 cm and CE = 1 cm, determine the length of AF. Solution Since ABCD is a rectangle, BC = AD = 12. Then BE = BC + CE = = 13. F A B C D E Since BEDF is a quarter circle with centre B, BF = BD = BE = 13. Using the Pythagorean Theorem in right BCD, DC 2 = DB 2 BC 2 = = = 25 and DC = 5 (since DC > 0). Since ABCD is a rectangle, AB = DC = 5. Then AF = BF AB = 13 5 = 8 cm. Therefore, the length of AF is 8 cm.

16 A Problem C Shady Rectangles K B E F G H D J C A rectangle ABCD has a square AEF K of area 4 cm 2 and a square of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles.

17 Problem C and Solution Shady Rectangles Problem A rectangle ABCD has a square AEF K of area 4 cm 2 and a square GHCJ of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles. A K B E F G H D J C Solution Since square AEF K has area 4 cm 2, its side lengths must be AE = EF = 2 cm. Since square GHCJ has area 9 cm 2, its side lengths must be JC = CH = 3 cm. Since line segment AKB passes through the top of square AEF K and line segment EF GH passes through the bottom of the same square, AKB EF GH and it follows that AB = EH. Therefore, AB = EH = EF + F G + GH = = 10 cm. Also, EDJG is a rectangle so ED = GJ = HC = 3 cm. Then AD = AE + ED = = 5 cm. The area of the shaded rectangles can be calculated by finding the total area of rectangle ABCD and subtracting the area of the two squares that have been removed. Area of Shaded Rectangles = Area ABCD Area AEF K Area GHCJ the shaded area is 37 cm 2. = AB AD 4 9 = = 37 cm 2 It should be noted that we could also find the areas of each of the rectangles EDJG and KF HB and then determine the sum of the areas achieving the same result.

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19 Problem C A Bit of Shade Please A B 3 cm O In the diagram, AOB = 90 and the radius of the circle is OA = 3 cm. Determine the area of the shaded region correct to one decimal place.

20 Problem C and Solution A Bit of Shade Please Problem In the diagram, AOB = 90 and the radius of the circle is OA = 3 cm. Determine the area of the shaded region correct to one decimal place. A 3 cm O B Solution The triangle plus the shaded area cover 1 4 of the area of the circle. To find the shaded area we need to find the area of the triangle and subtract it from the area of 1 4 of a circle with radius 3 cm. Since AOB = 90, AO and OB meet at 90, we can use OA as the base and OB as the height in the formula for the area of a triangle, base height 2. Area AOB = (AO) (OB) 2 = (3) (3) (2) = 9 2 = 4.5 cm 2. To find the area of the quarter circle we will use the formula πr 2 and divide the result by 4. Area of Quarter Circle = π r 2 4 = π = π 9 4 = 2.25π cm 2. We can now determine the shaded area. Shaded area = Area of Quarter Circle Area of Triangle = 2.25π 4.5. = 2.6 cm 2. Therefore, the shaded area is approximately 2.6 cm 2.

21 Problem C Rectangles Abound A D B C Seven identical rectangles are arranged as shown in the diagram to form a large rectangle ABCD. If the area of rectangle ABCD is 560 cm 2, determine the dimensions of the smaller rectangles.

22 Problem Problem C and Solutions Rectangles Abound Seven identical rectangles are arranged as shown in the diagram to form a large rectangle ABCD. If the area of rectangle ABCD is 560 cm 2, determine the dimensions of the smaller rectangles. Solution 1 Let x be the width of one of the smaller identical rectangles, in cm. Five of the smaller rectangles are stacked on top of each other creating AB, so AB = x + x + x + x + x = 5x. Since ABCD is a rectangle, AB = CD = 5x. But CD is the length of the smaller rectangle. Therefore, the smaller rectangle is 5x cm by x cm. A D x x x x x B 5x The area of rectangle ABCD is the same as 7 times the area of one of the smaller rectangles. Area ABCD = 7 Area of one smaller rectangle 560 = 7 5x x 560 = 35 x 2 Dividing both sides by 35, we obtain x 2 = 16 and x = 4 follows. (x > 0 since x is the width of the smaller rectangle.) The width of the smaller rectangle is x = 4 cm and the length of the smaller rectangle is 5x = 5(4) = 20 cm. Therefore, the smaller rectangle is 20 cm long and 4 cm wide. x x 5x C

23 Solution 2 Let x be the width of one of the smaller identical rectangles, in cm. Five rectangles are stacked on top of each other creating AB, so AB = x + x + x + x + x = 5x. A x D x x x 5x x B 5x x x C But ABCD is a rectangle so AB = CD. The width of rectangle ABCD is CD = 5x. Now BC is made up of the length of the smaller rectangle plus two widths of the smaller rectangle. Therefore, BC = 5x + x + x = 7x and rectangle ABCD is 7x cm long and 5x cm wide. To find the area of ABCD we multiply the length BC by the width CD. Area ABCD = BC CD 560 = (7x) (5x) 560 = 7 5 (x) (x) 560 = 35 x 2 Dividing by 35, we obtain x 2 = 16 and x = 4 follows. (x > 0 since x is the width of the smaller rectangle.) The width of the smaller rectangle is x = 4 cm and the length of the smaller rectangle is 5x = 5(4) = 20 cm. Therefore, the smaller rectangle is 20 cm long and 4 cm wide.

24 Problem C Am I There Yet? Two towns are 120 km apart. Cara Van wants to drive from one town to the other in exactly one hour and twenty minutes. For the first 30 minutes, Cara drives at a constant rate of 70 km/h. At what constant rate, in km/h, must she drive for the remaining time if she is to accomplish her goal?

25 Problem C and Solution Am I There Yet? Problem Two towns are 120 km apart. Cara Van wants to drive from one town to the other in exactly one hour and twenty minutes. For the first 30 minutes, Cara drives at a constant rate of 70 km/h. At what constant rate, in km/h, must she drive for the remaining time if she is to accomplish her goal? Solution Representing the information in a diagram is often useful in helping to solve the problem. 1 hour 20 minutes 80 minutes 120 km 30 minutes 70 km / h 50 minutes?? km / h The total trip is one hour and twenty minutes or 80 minutes. For the first 30 minutes, Cara travels at a constant rate of 70 km/h. This means that in one hour (60 minutes) she would travel 70 km. Therefore, in half the time or 30 minutes she would travel half the distance or 70 2 = 35 km. So Cara must drive = 85 km in = 50 minutes. We need to determine the constant rate that Cara needs to drive to accomplish this. Cara needs to drive 85 km in 50 minutes. By dividing each term by 5, Cara needs to drive 85 5 = 17 km in 50 5 = 10 minutes. Multiplying each term by 6, Cara must drive 17 6 = 102 km in 10 6 = 60 minutes (1 hour). Therefore, Cara must drive 102 km/h to accomplish her goal of driving 120 km to her destination in one hour and twenty minutes.

26 Problem C Reach For The Top A rectangular storage tank has a square base with sides of length 4 m and height of 5 m. The tank is filled with water to a height of 2.5 m. A solid cube with sides 2 m is then thrown into the tank. Does the water level reach the top of the tank? If not, how far below the top of the tank does the water reach? 2 m 5 m 2 m 2 m 2.5 m 4 m 4 m

27 Problem C and Solution Reach For The Top Problem A rectangular storage tank has a square base with sides of length 4 m and height of 5 m. The tank is filled with water to a height of 2.5 m. A solid cube with sides 2 m is then thrown into the tank. Does the water level reach the top of the tank? If not, how far below the top of the tank does the water reach? Solution First calculate the volume of water in the tank using V olume = Length W idth Height. Volume of Water = = 40 m 3 The volume of the solid cube is = 8 m 3. The total volume of water plus solid cube is = 48 m 3. Let x represent the height of the water in the rectangular storage tank after the cube is thrown in. New Volume = Length Width Height 48 = 4 4 x 48 = 16 x x = 3 m The new water height is 3 m and the water is 5 3 = 2 m from the top of the tank.

28 A Problem C Shady Rectangles K B E F G H D J C A rectangle ABCD has a square AEF K of area 4 cm 2 and a square of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles.

29 Problem C and Solution Shady Rectangles Problem A rectangle ABCD has a square AEF K of area 4 cm 2 and a square GHCJ of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles. A K B E F G H D J C Solution Since square AEF K has area 4 cm 2, its side lengths must be AE = EF = 2 cm. Since square GHCJ has area 9 cm 2, its side lengths must be JC = CH = 3 cm. Since line segment AKB passes through the top of square AEF K and line segment EF GH passes through the bottom of the same square, AKB EF GH and it follows that AB = EH. Therefore, AB = EH = EF + F G + GH = = 10 cm. Also, EDJG is a rectangle so ED = GJ = HC = 3 cm. Then AD = AE + ED = = 5 cm. The area of the shaded rectangles can be calculated by finding the total area of rectangle ABCD and subtracting the area of the two squares that have been removed. Area of Shaded Rectangles = Area ABCD Area AEF K Area GHCJ the shaded area is 37 cm 2. = AB AD 4 9 = = 37 cm 2 It should be noted that we could also find the areas of each of the rectangles EDJG and KF HB and then determine the sum of the areas achieving the same result.

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31 Problem C A Bit of Shade Please A B 3 cm O In the diagram, AOB = 90 and the radius of the circle is OA = 3 cm. Determine the area of the shaded region correct to one decimal place.

32 Problem C and Solution A Bit of Shade Please Problem In the diagram, AOB = 90 and the radius of the circle is OA = 3 cm. Determine the area of the shaded region correct to one decimal place. A 3 cm O B Solution The triangle plus the shaded area cover 1 4 of the area of the circle. To find the shaded area we need to find the area of the triangle and subtract it from the area of 1 4 of a circle with radius 3 cm. Since AOB = 90, AO and OB meet at 90, we can use OA as the base and OB as the height in the formula for the area of a triangle, base height 2. Area AOB = (AO) (OB) 2 = (3) (3) (2) = 9 2 = 4.5 cm 2. To find the area of the quarter circle we will use the formula πr 2 and divide the result by 4. Area of Quarter Circle = π r 2 4 = π = π 9 4 = 2.25π cm 2. We can now determine the shaded area. Shaded area = Area of Quarter Circle Area of Triangle = 2.25π 4.5. = 2.6 cm 2. Therefore, the shaded area is approximately 2.6 cm 2.

33 Problem C This is Some Sum In the late 1700s, Gauss was asked to find the sum of the numbers from 1 to 100. Gauss quickly gave the answer He did this by looking at patterns. Instead of finding the sum of the numbers 1 to 100, can you find the sum of the digits of the numbers from 1 to 100? For example, the sum of the digits of the numbers from 1 to 14 is (1 + 0) + (1 + 1) + (1 + 2) + (1 + 3) + (1 + 4) = 60.

34 Problem Problem C and Solution This is Some Sum In the late 1700s, Gauss was asked to find the sum of the numbers from 1 to 100. Gauss quickly gave the answer He did this by looking at patterns. Instead of finding the sum of the numbers 1 to 100, can you find the sum of the digits of the numbers from 1 to 100? Solution (1) Each of the ten columns has a units digit that occurs ten times. Sum of ALL units digits = 10(1) + 10(2) + 10(3) (9) + 10(0) = 10( ) = 10(45) = 450 (2) Each of the ten columns has a ten s digit from 0 to 9. Sum of ALL tens digits = 10( ) = 10(45) = 450 (3) The number 100 is the only number with a hundreds digit. We need to add 1 to our final sum. (4) Now we add our results from (1), (2), and (3) to obtain the required sum. Sum of digits = Units digit sum + Tens digit sum + Hundreds Digit = = 901 Therefore the sum of the digits of the numbers from 1 to 100 is 901.

35 Problem C POWERful 8 3 means and equals 512 when expressed as an integer. When is expressed as an integer, what is its last digit?

36 Problem Problem C and Solution POWERful 8 3 means and equals 512 when expressed as an integer. When is expressed as an integer, what is its last digit? Solution Let s start by examining the last digit of various powers of = = = = = = = = Notice that the last digit repeats every four powers of 8. The pattern continues. 8 9 ends with 8, 8 10 ends with 4, 8 11 ends with 2, 8 12 ends with 6, and so on. Starting with the first power of 8, every four consecutive powers of 8 will have the last digit 8, 4, 2, and 6. We need to determine the number of complete cycles by dividing 2011 by = There are 502 complete cycles and 3 4 of another cycle = 2008 so is the last power of 8 in the 502 nd cycle and therefore ends with 6. To go 3 4 of the way into the next cycle tells us that the number ends with the third number in the pattern, namely 2. In fact, we know that ends with 8, ends with 4, ends with 2, and ends with 6 because they would be the numbers in the 503 rd complete cycle of the pattern. Therefore, ends with the digit 2.

37 Problem C A Time For Change A customer purchased some cough candies for $1.73 and paid for them with a two-dollar coin. The cashier made the correct change using only pennies, nickels, dimes and quarters. In how many different ways can the cashier make change?

38 Problem C and Solution A Time For Change Problem A customer purchased some cough candies for $1.73 and paid for them with a two-dollar coin. The cashier made the correct change using only pennies, nickels, dimes and quarters. In how many different ways can the cashier make change? Solution This is a good problem for applying a systematic approach. The amount of change required is $2 $1.73 = $0.27 or 27 cents. In order to get to 27 a minimum of 2 pennies are required. We can systematically look at all of the possibilities using pennies. The number of pennies given must end in a 2 or a 7 so that when this number is subtracted from 27 the result will be a number ending in 0 or 5. This result can then be achieved using nickels, dimes or quarters. Therefore, we can use 2, 7, 12, 17, 22 or 27 pennies. The following chart presents the possibilities. Number of Pennies Number of Nickels Number of Dimes Number of Quarters there are 13 different ways to make the correct change.

39 Problem C Chocolates or Cards? For Valentine s Day, in a class of 30 students, 12 students brought chocolates for the teacher, 17 brought the teacher a card, and 5 students did both. How many students did not bring the teacher chocolates or a card?

40 Problem C and Solution Chocolates or Cards? Problem For Valentine s Day, in a class of 30 students, 12 students brought chocolates for the teacher, 17 brought the teacher a card, and 5 students did both. How many students did not bring the teacher chocolates or a card? Solution Since 5 students did both and these students are included in the 12 who brought chocolates, then 12 5 or 7 students brought chocolates only. They did not bring a card for the teacher as well. Again, since 5 students did both and these students are included in the 17 who brought the teacher a card, then 17 5 or 12 students brought a card only. They did not bring chocolates for the teacher as well. Students will be in one of four possible groups: they brought both chocolates and a card, they brought chocolates only, they brought a card only, or they did not bring a card or chocolates. The number of students in each group added together will sum to the number of students in the class. Or we could subtract the sizes of the known groups from the class size to get the number of students who did neither. So, the number of students who did neither is the number of students in the class minus the number of students who brought both chocolates and a card minus the number of students who brought chocolates only minus the number of students who brought a card only. Therefore the number of students who did neither is equal to or 6. Six students did not bring their teacher chocolates or a card. These students will have to find some other way to show their fondness for their teacher.

41 Problem C Life is Not Fair A sum of money is to be divided among three children: Alex, Bogdan, and Chai. The money will be split as follows: (i) Alex receives $500 plus 1 5 of what then remains; (ii) Bogdan then receives $800 plus 1 4 of what then remains; and (iii) Chai then receives the rest, which is $900. How much is the original sum of money? Which child receives the most money?

42 Problem C and Solution Life is Not Fair Problem A sum of money is to be divided among three children: Alex, Bogdan, and Chai. The money will be split as follows: (i) Alex receives $500 plus 1 of what then remains; 5 (ii) Bogdan then receives $800 plus 1 of what then remains; and 4 (iii) Chai then receives the rest, which is $900. How much is the original sum of money? Which child receives the most money? Solution Start from Chai and work towards Alex. Bogdan received 1 4 of the remainder so what is left for Chai is 3 4 of the remainder. So $900 is 3 4 of the remainder. If 3 of the remainder is $900 then 1 of the remainder is $ So just after Bodgan received $800, there was $300 + $900 or $1 200 left. Therefore, before Bogdan got any money there was $ $800 or $ Bogdan received $800 + $300 = $ Alex received 1 of the remainder so what is left for Bogdan is 4 of the remainder. So $2 000 is 5 5 of the remainder. 4 5 If 4 of the remainder is $2 000 then 1 of the remainder is $ So just after Alex received $500, there was $500 + $2 000 or $2 500 left. Therefore, before Alex got any money there was $ $500 or $ Alex received $500 + $500 = $ Alex received $1 000, Bogdan received $1 100, and Chai received $900. the original sum of money was $3 000 and Bogdan received the most money.

43 Problem C The Search is On The digits 1, 2, 3, 4, and 5 are each used exactly once to create a five digit number abcde which satisfies the following two conditions: (i) the two digit number ab is divisible by 4; and (ii) the two digit number cd is divisible by 3. Find all five digit numbers, formed using each of the digits 1 to 5 exactly once, that satisfy both conditions.

44 Problem C and Solution The Search is On Problem The digits 1, 2, 3, 4, and 5 are each used exactly once to create a five digit number abcde which satisfies the following two conditions: the two digit number ab is divisible by 4; and the two digit number cd is divisible by 3. Find all five digit numbers, formed using each of the digits 1 to 5 exactly once, that satisfy both conditions. Solution The following two digit numbers are divisible by 4: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, and 52. From this list, 16, 20, 28, 36, 40 and 48 can be eliminated because they use a digit not permitted in the problem. The number 44 can be removed since a digit is repeated. The list stops at 52 since all multiples of 4 beyond this point use a digit that is not permitted in the problem. So the only two digit numbers that satisfy the first condition are 12, 24, 32 and 52. Therefore the five digit number looks like 12cde or 24cde or 32cde or 52cde. For a number to be divisible by three, the sum of its digits will be divisible by three. Since cd is divisible by three, the sum c + d must be divisible by three. Let s look at the four numbers, 12cde, 24cde, 32cde and 52cde, separately. If the number is 12cde, ab = 12 and there are only three possible digits left, 3, 4 and 5. If we use 3 and 4 for c and d, the sum is c + d = 7 which is not divisible by three. If we use 3 and 5 for c and d, the sum is c + d = 8 which is not divisible by three. If we use 4 and 5 for c and d, the sum is c + d = 9 which is divisible by three. But the order in which we use 4 and 5 does not matter for cd to be divisible by three. (The sum c + d would still be 9.) So cd = 45 or cd = 54. If ab = 12 and cd = 45 then e = 3 and one possible solution is abcde = If ab = 12 and cd = 54 then e = 3 and another possible solution is abcde = There are two numbers of the form 12cde that can be made which satisfy both conditions in the problem: and (The solution continues on the next page.)

45 If the number is 24cde, ab = 24 and there are only three possible digits left, 1, 3 and 5. If we use 1 and 3 for c and d, the sum is c + d = 4 which is not divisible by three. If we use 1 and 5 for c and d, the sum is c + d = 6 which is divisible by three. But the order in which we use 1 and 5 does not matter for cd to be divisible by three. So cd = 15 or cd = 51. If ab = 24 and cd = 15 then e = 3 and one possible solution is abcde = If ab = 24 and cd = 51 then e = 3 and another possible solution is abcde = If we use 3 and 5 for c and d, the sum is c + d = 8 which is not divisible by three. There are two numbers of the form 24cde that can be made which satisfy both conditions in the problem: and If the number is 32cde, ab = 32 and there are only three possible digits left, 1, 4 and 5. If we use 1 and 4 for c and d, the sum is c + d = 5 which is not divisible by three. If we use 1 and 5 for c and d, the sum is c + d = 6 which is divisible by three. But the order in which we use 1 and 5 does not matter for cd to be divisible by three. So cd = 15 or cd = 51. If ab = 32 and cd = 15 then e = 4 and one possible solution is abcde = If ab = 32 and cd = 51 then e = 4 and another possible solution is abcde = If we use 4 and 5 for c and d, the sum is c + d = 9 which is divisible by three. But the order in which we use 4 and 5 does not matter for cd to be divisible by three. So cd = 45 or cd = 54. If ab = 32 and cd = 45 then e = 1 and one possible solution is abcde = If ab = 32 and cd = 54 then e = 1 and another possible solution is abcde = There are four numbers of the form 32cde that can be made which satisfy both conditions in the problem: 32154, 32514, and If the number is 52cde, ab = 52 and there are only three possible digits left, 1, 3 and 4. If we use 1 and 3 for c and d, the sum is c + d = 4 which is not divisible by three. If we use 1 and 4 for c and d, the sum is c + d = 5 which is not divisible by three. If we use 3 and 4 for c and d, the sum is c + d = 7 which is not divisible by three. There are no numbers of the form 52cde that can be made which satisfy both conditions in the problem. Summarizing from each of the cases, there are 8 numbers, 12453, 12543, 24153, 24513, 32154, 32514, and 32541, which satisfy all of the conditions of the problem. Note: If a student were to try to solve this problem by listing all of the possible five digit numbers, there would be 120 numbers to check.

46 Problem C Pesky Products Three positive numbers exist such that the following is true: (i) the product of the first and second numbers equals the third number; (ii) the product of the second and third numbers is 180; and (iii) the second number is five times the third number. Determine the product of the three numbers.

47 Problem C and Solution Pesky Products Problem Three positive numbers exist such that the following is true: the product of the first and second numbers equals the third number; the product of the second and third numbers is 180; and the second number is five times the third number. Determine the product of the three numbers. Solution Let the three numbers be represented by a, b, and c. Since the product of the first and second numbers equals the third number, a b = c. We are looking for a b c = (a b) c = (c) c = c 2. So when we find c 2 we have found the required product a b c. We know that b c = 180 and b = 5 c so b c = 180 becomes (5 c) c = 180 or 5 c 2 = 180. Dividing by 5, we obtain c 2 = 36. This is exactly what we are looking for since a b c = c 2. Therefore, the product of the three numbers is 36. For those who need to know what the actual numbers are, we can proceed and find the three numbers. We know c 2 = 36, so c = 6 since c is a positive number. So b = 5 c = 5 (6) = 30. And finally, a b = c so a (30) = 6. Dividing by 30, we get a = 6 30 = 1 5 = 0.2. We can verify the product a b c = (0.2) (30) (6) = 6 6 = 36.

48 Problem C Where Are You? The integers greater than one are arranged as shown in the following chart. P Q R S T Row Row Row Row Row Row The pattern continues. Determine the exact position of the integer State the row number and the column letter (P, Q, R, S, T ).

49 Problem C and Solution Where Are You? Problem The integers greater than one are arranged as shown in the following chart. P Q R S T Row Row Row Row Row Row The pattern continues. Determine the exact position of the integer State the row number and the column letter (P, Q, R, S, T ). Solution Observe some of the patterns in the chart. Each row contains a multiple of 3 in either column Q or column S. To determine the row number, take the multiple of three and divide it by three. The outer numbers in column P or T have a remainder 1 when divided by 3. Numbers that are even and have a remainder 1 when divided by 3 are in column T. Numbers that are odd and have a remainder 1 when divided by 3 are in column P. If the largest number in a row is even, it is in column T. If the largest number in a row is odd, it is in column P. Every number in column R has a remainder 2 when divided by 3. When 2011 is divided by 3, there is a quotient of 670 and a remainder 1. So 2011 is in column P or T but since 2011 is odd, it is in column P. The 670 th multiple of 3, 2010, is in row 670, in column Q, to the right of In fact, row 670 will contain 2011 in column P, 2010, the 670 th multiple of 3, in column Q, and 2009 in column R.

50 Problem C Rectangles Abound A D B C Seven identical rectangles are arranged as shown in the diagram to form a large rectangle ABCD. If the area of rectangle ABCD is 560 cm 2, determine the dimensions of the smaller rectangles.

51 Problem Problem C and Solutions Rectangles Abound Seven identical rectangles are arranged as shown in the diagram to form a large rectangle ABCD. If the area of rectangle ABCD is 560 cm 2, determine the dimensions of the smaller rectangles. Solution 1 Let x be the width of one of the smaller identical rectangles, in cm. Five of the smaller rectangles are stacked on top of each other creating AB, so AB = x + x + x + x + x = 5x. Since ABCD is a rectangle, AB = CD = 5x. But CD is the length of the smaller rectangle. Therefore, the smaller rectangle is 5x cm by x cm. A D x x x x x B 5x The area of rectangle ABCD is the same as 7 times the area of one of the smaller rectangles. Area ABCD = 7 Area of one smaller rectangle 560 = 7 5x x 560 = 35 x 2 Dividing both sides by 35, we obtain x 2 = 16 and x = 4 follows. (x > 0 since x is the width of the smaller rectangle.) The width of the smaller rectangle is x = 4 cm and the length of the smaller rectangle is 5x = 5(4) = 20 cm. Therefore, the smaller rectangle is 20 cm long and 4 cm wide. x x 5x C

52 Solution 2 Let x be the width of one of the smaller identical rectangles, in cm. Five rectangles are stacked on top of each other creating AB, so AB = x + x + x + x + x = 5x. A x D x x x 5x x B 5x x x C But ABCD is a rectangle so AB = CD. The width of rectangle ABCD is CD = 5x. Now BC is made up of the length of the smaller rectangle plus two widths of the smaller rectangle. Therefore, BC = 5x + x + x = 7x and rectangle ABCD is 7x cm long and 5x cm wide. To find the area of ABCD we multiply the length BC by the width CD. Area ABCD = BC CD 560 = (7x) (5x) 560 = 7 5 (x) (x) 560 = 35 x 2 Dividing by 35, we obtain x 2 = 16 and x = 4 follows. (x > 0 since x is the width of the smaller rectangle.) The width of the smaller rectangle is x = 4 cm and the length of the smaller rectangle is 5x = 5(4) = 20 cm. Therefore, the smaller rectangle is 20 cm long and 4 cm wide.

53 Problem C Better Than Average Stu Dent has six marks on his report card. His overall average is 75%. But after looking more closely at his marks, Stu discovered that his mathematics mark was incorrectly recorded as a 39% instead of his actual mark of 93%. Determine Stu Dent s correct report card average.

54 Problem C and Solution Better Than Average Problem Stu Dent has six marks on his report card. His overall average is 75%. But after looking more closely at his marks, Stu discovered that his mathematics mark was incorrectly recorded as a 39% instead of his actual mark of 93%. Determine Stu Dent s correct report card average. Solution To calculate an average we add the six marks and divide by 6. sum of six marks 6 = 75 To then obtain the sum of the six marks we would multiply the average by 6. sum of six marks = 6 75 = 450 But this sum includes the wrong mathematics mark of 39%. So we need to adjust the sum by subtracting the wrong mark and adding the corrected mark. correct sum of six marks = = 504 We can now obtain Stu s corrected average by dividing the corrected sum by 6. correct average = correct sum of six marks 6 = = 84 Stu s corrected average is 84%. His day is now much better than average!

55 Problem C More Change Is Upon Us Notta Looney has 45 coins with a total value of $1.95. The coins are nickels, dimes and pennies. She has twice as many pennies as dimes and five more nickels than dimes. How many nickels does Notta have?

56 Problem C and Solution More Change Is Upon Us Problem Notta Looney has 45 coins with a total value of $1.95. The coins are nickels, dimes and pennies. She has twice as many pennies as dimes and five more nickels than dimes. How many nickels does Notta have? Solution At first, this problem appears similar to the problem in week 5, A Time For Change. In that problem we were trying to find the number of different possible combinations. A systematic approach worked very well. But in this problem we are looking for a single solution. A systematic approach is far less suitable. This problem is solved more efficiently using an algebraic approach. Let d represent the number of dimes Notta has. Since she has twice as many pennies as dimes, Notta has 2d dimes. Since she has five more nickels than dimes, she has d + 5 dimes. Since the total number of coins is 45, the number of dimes plus the number of pennies plus the number of nickels totals 45 coins. Therefore, d + 2d + d + 5 = 45. This simplifies to 4d + 5 = 45 or 4d = 40 or d = 10. From here 2d = 20 and d + 5 = 15. Therefore, Notta has 10 dimes, 20 pennies and 15 nickels. We can use the total value to check the accuracy of our solution. Number of Number of Number of Dimes Pennies Nickels Value of Value of Value of Total Dimes Pennies Nickels Value = $ = $ = $0.75 $1.95

57 Problem C Am I There Yet? Two towns are 120 km apart. Cara Van wants to drive from one town to the other in exactly one hour and twenty minutes. For the first 30 minutes, Cara drives at a constant rate of 70 km/h. At what constant rate, in km/h, must she drive for the remaining time if she is to accomplish her goal?

58 Problem C and Solution Am I There Yet? Problem Two towns are 120 km apart. Cara Van wants to drive from one town to the other in exactly one hour and twenty minutes. For the first 30 minutes, Cara drives at a constant rate of 70 km/h. At what constant rate, in km/h, must she drive for the remaining time if she is to accomplish her goal? Solution Representing the information in a diagram is often useful in helping to solve the problem. 1 hour 20 minutes 80 minutes 120 km 30 minutes 70 km / h 50 minutes?? km / h The total trip is one hour and twenty minutes or 80 minutes. For the first 30 minutes, Cara travels at a constant rate of 70 km/h. This means that in one hour (60 minutes) she would travel 70 km. Therefore, in half the time or 30 minutes she would travel half the distance or 70 2 = 35 km. So Cara must drive = 85 km in = 50 minutes. We need to determine the constant rate that Cara needs to drive to accomplish this. Cara needs to drive 85 km in 50 minutes. By dividing each term by 5, Cara needs to drive 85 5 = 17 km in 50 5 = 10 minutes. Multiplying each term by 6, Cara must drive 17 6 = 102 km in 10 6 = 60 minutes (1 hour). Therefore, Cara must drive 102 km/h to accomplish her goal of driving 120 km to her destination in one hour and twenty minutes.

59 Problem C Reach For The Top A rectangular storage tank has a square base with sides of length 4 m and height of 5 m. The tank is filled with water to a height of 2.5 m. A solid cube with sides 2 m is then thrown into the tank. Does the water level reach the top of the tank? If not, how far below the top of the tank does the water reach? 2 m 5 m 2 m 2 m 2.5 m 4 m 4 m

60 Problem C and Solution Reach For The Top Problem A rectangular storage tank has a square base with sides of length 4 m and height of 5 m. The tank is filled with water to a height of 2.5 m. A solid cube with sides 2 m is then thrown into the tank. Does the water level reach the top of the tank? If not, how far below the top of the tank does the water reach? Solution First calculate the volume of water in the tank using V olume = Length W idth Height. Volume of Water = = 40 m 3 The volume of the solid cube is = 8 m 3. The total volume of water plus solid cube is = 48 m 3. Let x represent the height of the water in the rectangular storage tank after the cube is thrown in. New Volume = Length Width Height 48 = 4 4 x 48 = 16 x x = 3 m The new water height is 3 m and the water is 5 3 = 2 m from the top of the tank.

61 A Problem C Shady Rectangles K B E F G H D J C A rectangle ABCD has a square AEF K of area 4 cm 2 and a square of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles.

62 Problem C and Solution Shady Rectangles Problem A rectangle ABCD has a square AEF K of area 4 cm 2 and a square GHCJ of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles. A K B E F G H D J C Solution Since square AEF K has area 4 cm 2, its side lengths must be AE = EF = 2 cm. Since square GHCJ has area 9 cm 2, its side lengths must be JC = CH = 3 cm. Since line segment AKB passes through the top of square AEF K and line segment EF GH passes through the bottom of the same square, AKB EF GH and it follows that AB = EH. Therefore, AB = EH = EF + F G + GH = = 10 cm. Also, EDJG is a rectangle so ED = GJ = HC = 3 cm. Then AD = AE + ED = = 5 cm. The area of the shaded rectangles can be calculated by finding the total area of rectangle ABCD and subtracting the area of the two squares that have been removed. Area of Shaded Rectangles = Area ABCD Area AEF K Area GHCJ the shaded area is 37 cm 2. = AB AD 4 9 = = 37 cm 2 It should be noted that we could also find the areas of each of the rectangles EDJG and KF HB and then determine the sum of the areas achieving the same result.

63 Problem C Made To Order Math A fast food restaurant sells three items: pitas, baked potato chips and bottled water. The price of an item does not change regardless of the quantity purchased. Purchasing a pita, a bag of baked potato chips and a bottled water costs $6.00. For $9.50 you can buy two pitas, a bag of baked potato chips and a bottled water. A pita and a bottled water can be purchased for $4.70. Cyril Dorder, known by his friends as Cy, purchases three pitas, two bags of baked potato chips and a bottled water. What will lunch cost Cy Dorder?

64 Problem C and Solutions Made To Order Math Problem A fast food restaurant sells three items: pitas, baked potato chips and bottled water. The price of an item does not change regardless of the quantity purchased. Purchasing a pita, a bag of baked potato chips and a bottled water costs $6.00. For $9.50 you can buy two pitas, a bag of baked potato chips and a bottled water. A pita and a bottled water can be purchased for $4.70. Cyril Dorder, known by his friends as Cy, purchases three pitas, two bags of baked potato chips and a bottled water. What will lunch cost Cy Dorder? Solution 1 In this solution we will present a logical, non-algebraic approach to solving the problem. The subscriber is also encouraged to look at Solution 2, an algebraic approach to solving the problem. For $6.00 you get a pita, a bag of potato chips and a bottle of water. For $9.50 you get the same items plus a second pita. The difference in the two prices is the cost of one pita. Therefore, one pita costs $9.50 $6.00 = $3.50. For $4.70 you get a pita and a bottle of water and we know that a pita costs $3.50. The difference must be the cost of a bottle of water so a bottle of water costs $4.70 $3.50 = $1.20. Finally we know that a pita costs $3.50, a bottle of water costs $1.20 and all three items cost $6.00. The difference between the cost of all three items and the cost of two of the items must be the cost of the third item so the cost of a bag of potato chips is $6.00 $3.50 $1.20 = $1.30. Once we know the cost of the three items individually we can determine the cost of Cyril s order. Three pitas cost 3 $3.50 = $10.50, two bags of potato chips cost 2 $1.30 = $2.60, and a bottle of water costs $1.20. The total cost is the sum of the three totals, $ $ $1.20 = $ the cost of Cy Dorder s lunch is $14.30.

65 Solution 2 In this solution we will present an algebraic approach to solving the problem. Let p represent the cost of one pita. Let b represent the cost of one bag of baked potato chips. Let w represent the cost of one bottle of water. A pita, a bag of baked potato chips and a bottled water cost $6.00 so p + b + w = $6.00. (1) Two pitas, a bag of potato chips and a bottled water cost $9.50 so 2p + b + w = $9.50. (2) A pita and a bottled water can be purchased for $4.70 so p + w = $4.70. (3) We will use equations (1) and (2) to solve for p. 2p + b + w = $9.50 (2) p + b + w = $6.00 (1) Subtracting (1) from (2), we obtain p = $3.50. We will use equations (1) and (3) to solve for b. p + b + w = $6.00 (1) p + w = $4.70 (3) Subtracting (3) from (1), we obtain b = $1.30. We can now substitute p = $3.50 and b = $1.30 into equation (1) to find w. p + b + w = $6.00 (1) $ $ w = $6.00 $ w = $6.00 w = $6.00 $4.80 w = $1.20 Cy Dorder wants three pitas, two bags of baked potato chips and a bottle of water. This translates to the algebraic expression 3p + 2b + w which we need to evaluate when p = $3.50, b = $1.30 and w = $1.20. Evaluating, we obtain 3p + 2b + w = 3($3.50) + 2($1.30) + ($1.20) = $ $ $1.20 = $ the cost of Cy Dorder s lunch is $14.30.

66

67 Problem C This is Some Sum In the late 1700s, Gauss was asked to find the sum of the numbers from 1 to 100. Gauss quickly gave the answer He did this by looking at patterns. Instead of finding the sum of the numbers 1 to 100, can you find the sum of the digits of the numbers from 1 to 100? For example, the sum of the digits of the numbers from 1 to 14 is (1 + 0) + (1 + 1) + (1 + 2) + (1 + 3) + (1 + 4) = 60.

68 Problem Problem C and Solution This is Some Sum In the late 1700s, Gauss was asked to find the sum of the numbers from 1 to 100. Gauss quickly gave the answer He did this by looking at patterns. Instead of finding the sum of the numbers 1 to 100, can you find the sum of the digits of the numbers from 1 to 100? Solution (1) Each of the ten columns has a units digit that occurs ten times. Sum of ALL units digits = 10(1) + 10(2) + 10(3) (9) + 10(0) = 10( ) = 10(45) = 450 (2) Each of the ten columns has a ten s digit from 0 to 9. Sum of ALL tens digits = 10( ) = 10(45) = 450 (3) The number 100 is the only number with a hundreds digit. We need to add 1 to our final sum. (4) Now we add our results from (1), (2), and (3) to obtain the required sum. Sum of digits = Units digit sum + Tens digit sum + Hundreds Digit = = 901 Therefore the sum of the digits of the numbers from 1 to 100 is 901.

69 Problem C POWERful 8 3 means and equals 512 when expressed as an integer. When is expressed as an integer, what is its last digit?

70 Problem Problem C and Solution POWERful 8 3 means and equals 512 when expressed as an integer. When is expressed as an integer, what is its last digit? Solution Let s start by examining the last digit of various powers of = = = = = = = = Notice that the last digit repeats every four powers of 8. The pattern continues. 8 9 ends with 8, 8 10 ends with 4, 8 11 ends with 2, 8 12 ends with 6, and so on. Starting with the first power of 8, every four consecutive powers of 8 will have the last digit 8, 4, 2, and 6. We need to determine the number of complete cycles by dividing 2011 by = There are 502 complete cycles and 3 4 of another cycle = 2008 so is the last power of 8 in the 502 nd cycle and therefore ends with 6. To go 3 4 of the way into the next cycle tells us that the number ends with the third number in the pattern, namely 2. In fact, we know that ends with 8, ends with 4, ends with 2, and ends with 6 because they would be the numbers in the 503 rd complete cycle of the pattern. Therefore, ends with the digit 2.

71 Problem C The Search is On The digits 1, 2, 3, 4, and 5 are each used exactly once to create a five digit number abcde which satisfies the following two conditions: (i) the two digit number ab is divisible by 4; and (ii) the two digit number cd is divisible by 3. Find all five digit numbers, formed using each of the digits 1 to 5 exactly once, that satisfy both conditions.

72 Problem C and Solution The Search is On Problem The digits 1, 2, 3, 4, and 5 are each used exactly once to create a five digit number abcde which satisfies the following two conditions: the two digit number ab is divisible by 4; and the two digit number cd is divisible by 3. Find all five digit numbers, formed using each of the digits 1 to 5 exactly once, that satisfy both conditions. Solution The following two digit numbers are divisible by 4: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, and 52. From this list, 16, 20, 28, 36, 40 and 48 can be eliminated because they use a digit not permitted in the problem. The number 44 can be removed since a digit is repeated. The list stops at 52 since all multiples of 4 beyond this point use a digit that is not permitted in the problem. So the only two digit numbers that satisfy the first condition are 12, 24, 32 and 52. Therefore the five digit number looks like 12cde or 24cde or 32cde or 52cde. For a number to be divisible by three, the sum of its digits will be divisible by three. Since cd is divisible by three, the sum c + d must be divisible by three. Let s look at the four numbers, 12cde, 24cde, 32cde and 52cde, separately. If the number is 12cde, ab = 12 and there are only three possible digits left, 3, 4 and 5. If we use 3 and 4 for c and d, the sum is c + d = 7 which is not divisible by three. If we use 3 and 5 for c and d, the sum is c + d = 8 which is not divisible by three. If we use 4 and 5 for c and d, the sum is c + d = 9 which is divisible by three. But the order in which we use 4 and 5 does not matter for cd to be divisible by three. (The sum c + d would still be 9.) So cd = 45 or cd = 54. If ab = 12 and cd = 45 then e = 3 and one possible solution is abcde = If ab = 12 and cd = 54 then e = 3 and another possible solution is abcde = There are two numbers of the form 12cde that can be made which satisfy both conditions in the problem: and (The solution continues on the next page.)

73 If the number is 24cde, ab = 24 and there are only three possible digits left, 1, 3 and 5. If we use 1 and 3 for c and d, the sum is c + d = 4 which is not divisible by three. If we use 1 and 5 for c and d, the sum is c + d = 6 which is divisible by three. But the order in which we use 1 and 5 does not matter for cd to be divisible by three. So cd = 15 or cd = 51. If ab = 24 and cd = 15 then e = 3 and one possible solution is abcde = If ab = 24 and cd = 51 then e = 3 and another possible solution is abcde = If we use 3 and 5 for c and d, the sum is c + d = 8 which is not divisible by three. There are two numbers of the form 24cde that can be made which satisfy both conditions in the problem: and If the number is 32cde, ab = 32 and there are only three possible digits left, 1, 4 and 5. If we use 1 and 4 for c and d, the sum is c + d = 5 which is not divisible by three. If we use 1 and 5 for c and d, the sum is c + d = 6 which is divisible by three. But the order in which we use 1 and 5 does not matter for cd to be divisible by three. So cd = 15 or cd = 51. If ab = 32 and cd = 15 then e = 4 and one possible solution is abcde = If ab = 32 and cd = 51 then e = 4 and another possible solution is abcde = If we use 4 and 5 for c and d, the sum is c + d = 9 which is divisible by three. But the order in which we use 4 and 5 does not matter for cd to be divisible by three. So cd = 45 or cd = 54. If ab = 32 and cd = 45 then e = 1 and one possible solution is abcde = If ab = 32 and cd = 54 then e = 1 and another possible solution is abcde = There are four numbers of the form 32cde that can be made which satisfy both conditions in the problem: 32154, 32514, and If the number is 52cde, ab = 52 and there are only three possible digits left, 1, 3 and 4. If we use 1 and 3 for c and d, the sum is c + d = 4 which is not divisible by three. If we use 1 and 4 for c and d, the sum is c + d = 5 which is not divisible by three. If we use 3 and 4 for c and d, the sum is c + d = 7 which is not divisible by three. There are no numbers of the form 52cde that can be made which satisfy both conditions in the problem. Summarizing from each of the cases, there are 8 numbers, 12453, 12543, 24153, 24513, 32154, 32514, and 32541, which satisfy all of the conditions of the problem. Note: If a student were to try to solve this problem by listing all of the possible five digit numbers, there would be 120 numbers to check.

74 Problem C Pesky Products Three positive numbers exist such that the following is true: (i) the product of the first and second numbers equals the third number; (ii) the product of the second and third numbers is 180; and (iii) the second number is five times the third number. Determine the product of the three numbers.

75 Problem C and Solution Pesky Products Problem Three positive numbers exist such that the following is true: the product of the first and second numbers equals the third number; the product of the second and third numbers is 180; and the second number is five times the third number. Determine the product of the three numbers. Solution Let the three numbers be represented by a, b, and c. Since the product of the first and second numbers equals the third number, a b = c. We are looking for a b c = (a b) c = (c) c = c 2. So when we find c 2 we have found the required product a b c. We know that b c = 180 and b = 5 c so b c = 180 becomes (5 c) c = 180 or 5 c 2 = 180. Dividing by 5, we obtain c 2 = 36. This is exactly what we are looking for since a b c = c 2. Therefore, the product of the three numbers is 36. For those who need to know what the actual numbers are, we can proceed and find the three numbers. We know c 2 = 36, so c = 6 since c is a positive number. So b = 5 c = 5 (6) = 30. And finally, a b = c so a (30) = 6. Dividing by 30, we get a = 6 30 = 1 5 = 0.2. We can verify the product a b c = (0.2) (30) (6) = 6 6 = 36.

76 Problem C Where Are You? The integers greater than one are arranged as shown in the following chart. P Q R S T Row Row Row Row Row Row The pattern continues. Determine the exact position of the integer State the row number and the column letter (P, Q, R, S, T ).

77 Problem C and Solution Where Are You? Problem The integers greater than one are arranged as shown in the following chart. P Q R S T Row Row Row Row Row Row The pattern continues. Determine the exact position of the integer State the row number and the column letter (P, Q, R, S, T ). Solution Observe some of the patterns in the chart. Each row contains a multiple of 3 in either column Q or column S. To determine the row number, take the multiple of three and divide it by three. The outer numbers in column P or T have a remainder 1 when divided by 3. Numbers that are even and have a remainder 1 when divided by 3 are in column T. Numbers that are odd and have a remainder 1 when divided by 3 are in column P. If the largest number in a row is even, it is in column T. If the largest number in a row is odd, it is in column P. Every number in column R has a remainder 2 when divided by 3. When 2011 is divided by 3, there is a quotient of 670 and a remainder 1. So 2011 is in column P or T but since 2011 is odd, it is in column P. The 670 th multiple of 3, 2010, is in row 670, in column Q, to the right of In fact, row 670 will contain 2011 in column P, 2010, the 670 th multiple of 3, in column Q, and 2009 in column R.

78 Problem C Rectangles Abound A D B C Seven identical rectangles are arranged as shown in the diagram to form a large rectangle ABCD. If the area of rectangle ABCD is 560 cm 2, determine the dimensions of the smaller rectangles.

79 Problem Problem C and Solutions Rectangles Abound Seven identical rectangles are arranged as shown in the diagram to form a large rectangle ABCD. If the area of rectangle ABCD is 560 cm 2, determine the dimensions of the smaller rectangles. Solution 1 Let x be the width of one of the smaller identical rectangles, in cm. Five of the smaller rectangles are stacked on top of each other creating AB, so AB = x + x + x + x + x = 5x. Since ABCD is a rectangle, AB = CD = 5x. But CD is the length of the smaller rectangle. Therefore, the smaller rectangle is 5x cm by x cm. A D x x x x x B 5x The area of rectangle ABCD is the same as 7 times the area of one of the smaller rectangles. Area ABCD = 7 Area of one smaller rectangle 560 = 7 5x x 560 = 35 x 2 Dividing both sides by 35, we obtain x 2 = 16 and x = 4 follows. (x > 0 since x is the width of the smaller rectangle.) The width of the smaller rectangle is x = 4 cm and the length of the smaller rectangle is 5x = 5(4) = 20 cm. Therefore, the smaller rectangle is 20 cm long and 4 cm wide. x x 5x C

80 Solution 2 Let x be the width of one of the smaller identical rectangles, in cm. Five rectangles are stacked on top of each other creating AB, so AB = x + x + x + x + x = 5x. A x D x x x 5x x B 5x x x C But ABCD is a rectangle so AB = CD. The width of rectangle ABCD is CD = 5x. Now BC is made up of the length of the smaller rectangle plus two widths of the smaller rectangle. Therefore, BC = 5x + x + x = 7x and rectangle ABCD is 7x cm long and 5x cm wide. To find the area of ABCD we multiply the length BC by the width CD. Area ABCD = BC CD 560 = (7x) (5x) 560 = 7 5 (x) (x) 560 = 35 x 2 Dividing by 35, we obtain x 2 = 16 and x = 4 follows. (x > 0 since x is the width of the smaller rectangle.) The width of the smaller rectangle is x = 4 cm and the length of the smaller rectangle is 5x = 5(4) = 20 cm. Therefore, the smaller rectangle is 20 cm long and 4 cm wide.

81 Problem C More Change Is Upon Us Notta Looney has 45 coins with a total value of $1.95. The coins are nickels, dimes and pennies. She has twice as many pennies as dimes and five more nickels than dimes. How many nickels does Notta have?

82 Problem C and Solution More Change Is Upon Us Problem Notta Looney has 45 coins with a total value of $1.95. The coins are nickels, dimes and pennies. She has twice as many pennies as dimes and five more nickels than dimes. How many nickels does Notta have? Solution At first, this problem appears similar to the problem in week 5, A Time For Change. In that problem we were trying to find the number of different possible combinations. A systematic approach worked very well. But in this problem we are looking for a single solution. A systematic approach is far less suitable. This problem is solved more efficiently using an algebraic approach. Let d represent the number of dimes Notta has. Since she has twice as many pennies as dimes, Notta has 2d dimes. Since she has five more nickels than dimes, she has d + 5 dimes. Since the total number of coins is 45, the number of dimes plus the number of pennies plus the number of nickels totals 45 coins. Therefore, d + 2d + d + 5 = 45. This simplifies to 4d + 5 = 45 or 4d = 40 or d = 10. From here 2d = 20 and d + 5 = 15. Therefore, Notta has 10 dimes, 20 pennies and 15 nickels. We can use the total value to check the accuracy of our solution. Number of Number of Number of Dimes Pennies Nickels Value of Value of Value of Total Dimes Pennies Nickels Value = $ = $ = $0.75 $1.95

83 Problem C Reach For The Top A rectangular storage tank has a square base with sides of length 4 m and height of 5 m. The tank is filled with water to a height of 2.5 m. A solid cube with sides 2 m is then thrown into the tank. Does the water level reach the top of the tank? If not, how far below the top of the tank does the water reach? 2 m 5 m 2 m 2 m 2.5 m 4 m 4 m

84 Problem C and Solution Reach For The Top Problem A rectangular storage tank has a square base with sides of length 4 m and height of 5 m. The tank is filled with water to a height of 2.5 m. A solid cube with sides 2 m is then thrown into the tank. Does the water level reach the top of the tank? If not, how far below the top of the tank does the water reach? Solution First calculate the volume of water in the tank using V olume = Length W idth Height. Volume of Water = = 40 m 3 The volume of the solid cube is = 8 m 3. The total volume of water plus solid cube is = 48 m 3. Let x represent the height of the water in the rectangular storage tank after the cube is thrown in. New Volume = Length Width Height 48 = 4 4 x 48 = 16 x x = 3 m The new water height is 3 m and the water is 5 3 = 2 m from the top of the tank.

85 Problem C Some Wet Weather Math? Raynor Schein has a bag that contains three black marbles, five gold marbles, two purple marbles, and six red marbles. During the course of cleaning his room one wet spring day, Raynor finds some white marbles and adds them to the bag. Raynor tells his friend April Showers that if she now draws a marble at random from the bag, the probability of it being black or gold is 2 7. How many white marbles did Raynor Schein add to the bag?

86 Problem C and Solutions Some Wet Weather Math? Problem Raynor Schein has a bag that contains three black marbles, five gold marbles, two purple marbles, and six red marbles. During the course of cleaning his room one wet spring day, Raynor finds some white marbles and adds them to the bag. Raynor tells his friend April Showers that if she now draws a marble at random from the bag, the probability of it being black or gold is 2. How many white marbles did Raynor Schein add to the bag? 7 Solution 1 At present there are = 16 marbles in the bag. Since after adding some marbles to the bag the probability of picking a black or gold marble is 2, this implies that the new total 7 number of marbles in the bag is a multiple of 7 and this multiple must be greater than 16. Therefore there are possibly 21, 28, 35, 42, 49, 56, marbles in the bag. There are a total of = 8 black and gold marbles in the bag. If k is the total number of marbles in the bag after adding some white marbles, then 8 k = 2 7 but 2 7 = 8 28 so 8 k = 8 28 and k = 28 follows. Since there were 16 marbles in the bag and there are now 28 marbles in the bag, Raynor added = 12 white marbles to the bag. Therefore, Raynor added 12 white marbles to the bag. Solution 2 Let w be the number of white marbles added to the bag. There are now w = 16 + w marbles in the bag and = 8 black and gold marbles in the bag. We know that the number of black and gold marbles divided by the total number of marbles in the bag is 2 7 so w = 2 7. Cross-multiplying we get (8)(7) = (2)(16 + w) which simplifies to 56 = w and 24 = 2w. The result w = 12 follows. Therefore, Raynor added 12 white marbles to the bag.

87 A Problem C Shady Rectangles K B E F G H D J C A rectangle ABCD has a square AEF K of area 4 cm 2 and a square of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles.

88 Problem C and Solution Shady Rectangles Problem A rectangle ABCD has a square AEF K of area 4 cm 2 and a square GHCJ of area 9 cm 2 removed. If EF GH is a straight line segment and F G = 5 cm, determine the total area of the two shaded rectangles. A K B E F G H D J C Solution Since square AEF K has area 4 cm 2, its side lengths must be AE = EF = 2 cm. Since square GHCJ has area 9 cm 2, its side lengths must be JC = CH = 3 cm. Since line segment AKB passes through the top of square AEF K and line segment EF GH passes through the bottom of the same square, AKB EF GH and it follows that AB = EH. Therefore, AB = EH = EF + F G + GH = = 10 cm. Also, EDJG is a rectangle so ED = GJ = HC = 3 cm. Then AD = AE + ED = = 5 cm. The area of the shaded rectangles can be calculated by finding the total area of rectangle ABCD and subtracting the area of the two squares that have been removed. Area of Shaded Rectangles = Area ABCD Area AEF K Area GHCJ the shaded area is 37 cm 2. = AB AD 4 9 = = 37 cm 2 It should be noted that we could also find the areas of each of the rectangles EDJG and KF HB and then determine the sum of the areas achieving the same result.

89 Problem C Made To Order Math A fast food restaurant sells three items: pitas, baked potato chips and bottled water. The price of an item does not change regardless of the quantity purchased. Purchasing a pita, a bag of baked potato chips and a bottled water costs $6.00. For $9.50 you can buy two pitas, a bag of baked potato chips and a bottled water. A pita and a bottled water can be purchased for $4.70. Cyril Dorder, known by his friends as Cy, purchases three pitas, two bags of baked potato chips and a bottled water. What will lunch cost Cy Dorder?